Question

At the instant when the current in an inductor is increasing at a rate of 0.0740 A/s, the magnitude of the self-induced emf is 0.0130 V.

a) What is the inductance of the inductor?
b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 A?

Answer 1

(a) Inductance will be 0.1756 Henry

(b) Flux through each turn will be
`3.16* 10^(-4)Wb`

Step-by-step explanation:

We have given the rate of change of current
`(di)/(dt)=0.0740A/sec`

Magnitude of self induced emf e = 0.0130 volt

(a) We know that in inductor self induced emf is given by

`e=L(di)/(dt)`

So
`0.0130=L* 0.0740`

L = 0.1756 Henry

Number of turns N = 400

Now magnetic flux through each turns is given by
`\Phi =(Li)/(N)=(0.1756* 0.720)/(400)=3.16* 10^(-4)Wb`